Angle between two straight lines Questions in English

(B) The line $2x + y + 10 = 0$ intersects the $x$-axis at $A(-5, 0)$ and the $y$-axis at $B(0, -10)$.
Let the points dividing $AB$ in the ratio $1:2:2$ be $P$ and $Q$. The total ratio is $1+2+2 = 5$.
The coordinates of $P$ are $\left( \frac{2(-5) + 3(0)}{5}, \frac{2(0) + 3(-10)}{5} \right) = (-2, -6)$.
The coordinates of $Q$ are $\left( \frac{4(-5) + 1(0)}{5}, \frac{4(0) + 1(-10)}{5} \right) = (-4, -2)$.
The slopes of lines $MP$ and $MQ$ are $m_1 = \frac{-6 - (-8)}{-2 - (-6)} = \frac{2}{4} = \frac{1}{2}$ and $m_2 = \frac{-2 - (-8)}{-4 - (-6)} = \frac{6}{2} = 3$.
The angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| = \left| \frac{3 - 0.5}{1 + 3(0.5)} \right| = \left| \frac{2.5}{2.5} \right| = 1$.
Thus,$\theta = \tan^{-1}(1) = \pi / 4$.

(C) Given curves are $y = 2x$ and $x^2 - xy + 2y^2 = 28$.
To find the intersection points,substitute $y = 2x$ into the second equation:
$x^2 - x(2x) + 2(2x)^2 = 28$
$x^2 - 2x^2 + 8x^2 = 28$
$7x^2 = 28 \implies x^2 = 4 \implies x = \pm 2$.
If $x = 2$,$y = 4$. If $x = -2$,$y = -4$. The intersection points are $(2, 4)$ and $(-2, -4)$.
For the first curve $y = 2x$,the slope $m_1 = \frac{dy}{dx} = 2$.
For the second curve $x^2 - xy + 2y^2 = 28$,differentiate with respect to $x$:
$2x - (y + x\frac{dy}{dx}) + 4y\frac{dy}{dx} = 0$
$\frac{dy}{dx}(4y - x) = y - 2x \implies \frac{dy}{dx} = \frac{y - 2x}{4y - x}$.
At $(2, 4)$,$m_2 = \frac{4 - 2(2)}{4(4) - 2} = \frac{0}{14} = 0$.
At $(-2, -4)$,$m_2 = \frac{-4 - 2(-2)}{4(-4) - (-2)} = \frac{0}{-14} = 0$.
The tangent of the angle $\theta$ between the curves is given by $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.
$\tan \theta = |\frac{2 - 0}{1 + 2(0)}| = |\frac{2}{1}| = 2$.

(A) The equation of the first line is $\alpha^2x + \alpha y = 9$,which can be written as $\alpha y = -\alpha^2x + 9$,or $y = -\alpha x + \frac{9}{\alpha}$. The slope $m_1$ is $-\alpha$.
The equation of the second line is $3x + 2y = 5$,which can be written as $2y = -3x + 5$,or $y = -\frac{3}{2}x + \frac{5}{2}$. The slope $m_2$ is $-\frac{3}{2}$.
Since the lines are perpendicular,the product of their slopes must be $-1$,so $m_1 \times m_2 = -1$.
Substituting the values,$(-\alpha) \times (-\frac{3}{2}) = -1$.
$\frac{3}{2}\alpha = -1$.
$\alpha = -\frac{2}{3}$.

(C) The given line is $\sqrt{3}x + y = 1$,which can be written as $y = -\sqrt{3}x + 1$. The slope of this line is $m_2 = -\sqrt{3}$.
Let the slope of line $L$ be $m$. The angle between the lines is $60^o$,so $\tan(60^o) = |\frac{m - m_2}{1 + m \cdot m_2}|$.
$\sqrt{3} = |\frac{m - (-\sqrt{3})}{1 + m(-\sqrt{3})}| = |\frac{m + \sqrt{3}}{1 - \sqrt{3}m}|$.
Squaring both sides: $3 = \frac{(m + \sqrt{3})^2}{(1 - \sqrt{3}m)^2} \Rightarrow 3(1 - 2\sqrt{3}m + 3m^2) = m^2 + 2\sqrt{3}m + 3$.
$3 - 6\sqrt{3}m + 9m^2 = m^2 + 2\sqrt{3}m + 3 \Rightarrow 8m^2 - 8\sqrt{3}m = 0$.
$8m(m - \sqrt{3}) = 0$,so $m = 0$ or $m = \sqrt{3}$.
If $m = 0$,the line is $y + 2 = 0(x - 3) \Rightarrow y + 2 = 0$,which is parallel to the $x$-axis and does not intersect it (unless it is the $x$-axis itself,which it is not).
If $m = \sqrt{3}$,the line is $y - (-2) = \sqrt{3}(x - 3) \Rightarrow y + 2 = \sqrt{3}x - 3\sqrt{3}$.
Rearranging gives $y - \sqrt{3}x + 2 + 3\sqrt{3} = 0$.

(A) The angle $\theta$ between two lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ is given by $\tan \theta = \left| \frac{a_1b_2 - a_2b_1}{a_1a_2 + b_1b_2} \right|$.
For the lines $2x + 3y + c_1 = 0$ and $-x + 5y + c_2 = 0$,the angle $\theta_1$ satisfies $\tan \theta_1 = \left| \frac{(2)(5) - (3)(-1)}{(2)(-1) + (3)(5)} \right| = \left| \frac{10 + 3}{-2 + 15} \right| = \frac{13}{13} = 1$.
Similarly,for the lines $2x + 3y + c_1 = 0$ and $-x + 5y + c_3 = 0$,the angle $\theta_2$ satisfies $\tan \theta_2 = \left| \frac{(2)(5) - (3)(-1)}{(2)(-1) + (3)(5)} \right| = 1$.
Since $\tan \theta_1 = \tan \theta_2 = 1$,we have $\theta_1 = \theta_2$ for any real values of $c_2$ and $c_3$.
Thus,Statement-$2$ is true.
Since Statement-$2$ is true,Statement-$1$ is also true,and Statement-$2$ provides the logical basis for Statement-$1$.

(D) The slopes of the lines are:
$m_1 = 1$ (for $L_1$)
$m_2 = -1$ (for $L_2$)
$m_3 = -1$ (for $L_3$)
$m_4 = 1$ (for $L_4$)
$1$. Check perpendicularity:
$m_1 \times m_2 = 1 \times (-1) = -1$,so $L_1 \perp L_2$.
$m_1 \times m_3 = 1 \times (-1) = -1$,so $L_1 \perp L_3$.
$2$. Check parallelism:
$m_2 = m_3 = -1$,so $L_2 || L_3$.
$3$. Check intersection:
Since $m_2 \neq m_4$ $(-1 \neq 1)$,$L_2$ and $L_4$ are not parallel and therefore intersect.
Comparing with the options,$L_1 \perp L_2$,$L_1 \perp L_3$,and $L_2$ intersects $L_4$ is the correct set of statements.

(D) The slope of the given line $2x - 3y + 17 = 0$ is $m_1 = \frac{-2}{-3} = \frac{2}{3}$.
Since the lines are perpendicular,the product of their slopes is $-1$. Let $m_2$ be the slope of the line passing through $(7, 17)$ and $(15, \beta)$.
$m_2 = \frac{\beta - 17}{15 - 7} = \frac{\beta - 17}{8}$.
Since $m_1 \times m_2 = -1$,we have $\frac{2}{3} \times \frac{\beta - 17}{8} = -1$.
$\frac{\beta - 17}{12} = -1$.
$\beta - 17 = -12$.
$\beta = 17 - 12 = 5$.

(N/A) We know that the acute angle $\theta$ between two lines with slopes $m_{1}$ and $m_{2}$ is given by $\tan \theta = \left| \frac{m_{2} - m_{1}}{1 + m_{1}m_{2}} \right| \dots (1)$
Let $m_{1} = \frac{1}{2},$ $m_{2} = m,$ and $\theta = \frac{\pi}{4}.$
Now,substituting these values in $(1),$ we get
$\tan \frac{\pi}{4} = \left| \frac{m - \frac{1}{2}}{1 + \frac{1}{2}m} \right| \implies 1 = \left| \frac{m - \frac{1}{2}}{1 + \frac{1}{2}m} \right|.$
This gives two cases:
Case $1: \frac{m - \frac{1}{2}}{1 + \frac{1}{2}m} = 1 \implies m - \frac{1}{2} = 1 + \frac{1}{2}m \implies \frac{1}{2}m = \frac{3}{2} \implies m = 3.$
Case $2: \frac{m - \frac{1}{2}}{1 + \frac{1}{2}m} = -1 \implies m - \frac{1}{2} = -1 - \frac{1}{2}m \implies \frac{3}{2}m = -\frac{1}{2} \implies m = -\frac{1}{3}.$
Hence,the slope of the other line is $3$ or $-\frac{1}{3}.$

(N/A) Let $m_{1}$ and $m_{2}$ be the slopes of the two lines such that $m_{1} = 2m_{2}$.
We know that if $\theta$ is the angle between two lines with slopes $m_{1}$ and $m_{2},$ then $\tan \theta = \left| \frac{m_{1} - m_{2}}{1 + m_{1}m_{2}} \right|$.
Given $\tan \theta = \frac{1}{3},$ we have $\frac{1}{3} = \left| \frac{2m_{2} - m_{2}}{1 + (2m_{2})m_{2}} \right| = \left| \frac{m_{2}}{1 + 2m_{2}^{2}} \right|$.
This implies $\frac{m_{2}}{1 + 2m_{2}^{2}} = \frac{1}{3}$ or $\frac{m_{2}}{1 + 2m_{2}^{2}} = -\frac{1}{3}$.
Case $I$: $\frac{m_{2}}{1 + 2m_{2}^{2}} = \frac{1}{3}$ $\Rightarrow 2m_{2}^{2} - 3m_{2} + 1 = 0$ $\Rightarrow (2m_{2} - 1)(m_{2} - 1) = 0$.
So,$m_{2} = \frac{1}{2}$ (giving $m_{1} = 1$) or $m_{2} = 1$ (giving $m_{1} = 2$).
Case $II$: $\frac{m_{2}}{1 + 2m_{2}^{2}} = -\frac{1}{3}$ $\Rightarrow 2m_{2}^{2} + 3m_{2} + 1 = 0$ $\Rightarrow (2m_{2} + 1)(m_{2} + 1) = 0$.
So,$m_{2} = -\frac{1}{2}$ (giving $m_{1} = -1$) or $m_{2} = -1$ (giving $m_{1} = -2$).
Thus,the possible pairs of slopes are $(1, \frac{1}{2}), (2, 1), (-1, -\frac{1}{2}), (-2, -1)$.

(A) The given lines are:
$y = \sqrt{3}x + 5$ ..... $(1)$
$\sqrt{3}y = x - 6 \implies y = \frac{1}{\sqrt{3}}x - 2\sqrt{3}$ ..... $(2)$
Comparing with $y = mx + c$,the slopes are $m_1 = \sqrt{3}$ and $m_2 = \frac{1}{\sqrt{3}}$.
The angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$.
Substituting the values: $\tan \theta = \left| \frac{\frac{1}{\sqrt{3}} - \sqrt{3}}{1 + (\sqrt{3})(\frac{1}{\sqrt{3}})} \right| = \left| \frac{\frac{1-3}{\sqrt{3}}}{1+1} \right| = \left| \frac{-2}{2\sqrt{3}} \right| = \frac{1}{\sqrt{3}}$.
Thus,$\theta = 30^{\circ}$.

The given lines can be written in slope-intercept form $y = mx + c$ as:
$y = -\frac{a_{1}}{b_{1}}x - \frac{c_{1}}{b_{1}}$ ..... $(1)$
$y = -\frac{a_{2}}{b_{2}}x - \frac{c_{2}}{b_{2}}$ ..... $(2)$
The slopes of lines $(1)$ and $(2)$ are $m_{1} = -\frac{a_{1}}{b_{1}}$ and $m_{2} = -\frac{a_{2}}{b_{2}}$ respectively.
Two lines are perpendicular if the product of their slopes is $-1$,i.e.,$m_{1} \cdot m_{2} = -1$.
Substituting the values of $m_{1}$ and $m_{2}$:
$(-\frac{a_{1}}{b_{1}}) \cdot (-\frac{a_{2}}{b_{2}}) = -1$
$\frac{a_{1}a_{2}}{b_{1}b_{2}} = -1$
$a_{1}a_{2} = -b_{1}b_{2}$
$a_{1}a_{2} + b_{1}b_{2} = 0$
Thus,the lines are perpendicular if $a_{1}a_{2} + b_{1}b_{2} = 0$.

(A) The given lines are $\sqrt{3}x + y = 1$ and $x + \sqrt{3}y = 1$.
Converting to slope-intercept form $y = mx + c$:
Line $(1): y = -\sqrt{3}x + 1$,so $m_1 = -\sqrt{3}$.
Line $(2): \sqrt{3}y = -x + 1 \implies y = -\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}$,so $m_2 = -\frac{1}{\sqrt{3}}$.
The acute angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values: $\tan \theta = \left| \frac{-\sqrt{3} - (-1/\sqrt{3})}{1 + (-\sqrt{3})(-1/\sqrt{3})} \right| = \left| \frac{(-3 + 1)/\sqrt{3}}{1 + 1} \right| = \left| \frac{-2}{2\sqrt{3}} \right| = \frac{1}{\sqrt{3}}$.
Thus,$\theta = 30^{\circ}$.
The obtuse angle is $180^{\circ} - 30^{\circ} = 150^{\circ}$.
Therefore,the angles between the lines are $30^{\circ}$ and $150^{\circ}$.

(A) The slope of the line passing through points $(h, 3)$ and $(4, 1)$ is $m_1 = \frac{1 - 3}{4 - h} = \frac{-2}{4 - h}$.
The slope of the line $7x - 9y - 19 = 0$ is $m_2 = \frac{7}{9}$.
Since the two lines are perpendicular,their product of slopes must be $-1$,so $m_1 \times m_2 = -1$.
$\left(\frac{-2}{4 - h}\right) \times \left(\frac{7}{9}\right) = -1$.
$\frac{-14}{36 - 9h} = -1$.
$14 = 36 - 9h$.
$9h = 36 - 14 = 22$.
$h = \frac{22}{9}$.

(N/A) Let the slope of the first line be $m_{1} = 2$ and the slope of the other line be $m_{2}$.
The angle $\theta$ between the two lines is given by $\tan \theta = \left| \frac{m_{1} - m_{2}}{1 + m_{1}m_{2}} \right|$.
Substituting $\theta = 60^{\circ}$ and $m_{1} = 2$:
$\tan 60^{\circ} = \left| \frac{2 - m_{2}}{1 + 2m_{2}} \right|$ $\Rightarrow \sqrt{3} = \left| \frac{2 - m_{2}}{1 + 2m_{2}} \right|$.
This gives two cases:
Case $I$: $\frac{2 - m_{2}}{1 + 2m_{2}} = \sqrt{3}$ $\Rightarrow 2 - m_{2} = \sqrt{3} + 2\sqrt{3}m_{2}$ $\Rightarrow m_{2}(1 + 2\sqrt{3}) = 2 - \sqrt{3}$ $\Rightarrow m_{2} = \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}}$.
The equation of the line passing through $(2,3)$ with this slope is $(y - 3) = \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}}(x - 2)$.
Simplifying,$(1 + 2\sqrt{3})y - 3(1 + 2\sqrt{3}) = (2 - \sqrt{3})x - 2(2 - \sqrt{3})$ $\Rightarrow (2 - \sqrt{3})x - (1 + 2\sqrt{3})y + (3 + 6\sqrt{3} - 4 + 2\sqrt{3}) = 0$ $\Rightarrow (2 - \sqrt{3})x - (1 + 2\sqrt{3})y + (8\sqrt{3} - 1) = 0$.
Case $II$: $\frac{2 - m_{2}}{1 + 2m_{2}} = -\sqrt{3}$ $\Rightarrow 2 - m_{2} = -\sqrt{3} - 2\sqrt{3}m_{2}$ $\Rightarrow m_{2}(2\sqrt{3} - 1) = -2 - \sqrt{3}$ $\Rightarrow m_{2} = -\frac{2 + \sqrt{3}}{2\sqrt{3} - 1}$.
The equation of the line passing through $(2,3)$ with this slope is $(y - 3) = -\frac{2 + \sqrt{3}}{2\sqrt{3} - 1}(x - 2)$.
Simplifying,$(2\sqrt{3} - 1)y - 3(2\sqrt{3} - 1) = -(2 + \sqrt{3})x + 2(2 + \sqrt{3})$ $\Rightarrow (2 + \sqrt{3})x + (2\sqrt{3} - 1)y - (4 + 2\sqrt{3} + 6\sqrt{3} - 3) = 0$ $\Rightarrow (2 + \sqrt{3})x + (2\sqrt{3} - 1)y - (1 + 8\sqrt{3}) = 0$.

Let the slope of the required line be $m_{1}$.
The given line is $x-2y=3$,which can be written as $y=\frac{1}{2}x-\frac{3}{2}$.
Thus,the slope of the given line is $m_{2}=\frac{1}{2}$.
The angle $\theta$ between the two lines is $45^{\circ}$. The formula for the angle between two lines with slopes $m_{1}$ and $m_{2}$ is $\tan \theta = \left| \frac{m_{2}-m_{1}}{1+m_{1}m_{2}} \right|$.
Substituting the values: $\tan 45^{\circ} = \left| \frac{\frac{1}{2}-m_{1}}{1+m_{1}(\frac{1}{2})} \right|$.
$1 = \left| \frac{1-2m_{1}}{2+m_{1}} \right|$.
This gives two cases:
Case $I$: $\frac{1-2m_{1}}{2+m_{1}} = 1$ $\Rightarrow 1-2m_{1} = 2+m_{1}$ $\Rightarrow 3m_{1} = -1$ $\Rightarrow m_{1} = -\frac{1}{3}$.
Case $II$: $\frac{1-2m_{1}}{2+m_{1}} = -1$ $\Rightarrow 1-2m_{1} = -2-m_{1}$ $\Rightarrow m_{1} = 3$.
For $m_{1} = 3$,the equation of the line passing through $(3,2)$ is $y-2 = 3(x-3)$ $\Rightarrow y-2 = 3x-9$ $\Rightarrow 3x-y = 7$.
For $m_{1} = -\frac{1}{3}$,the equation of the line passing through $(3,2)$ is $y-2 = -\frac{1}{3}(x-3)$ $\Rightarrow 3y-6 = -x+3$ $\Rightarrow x+3y = 9$.
Thus,the equations of the lines are $3x-y=7$ and $x+3y=9$.

Let the equation of the line passing through the origin be $y=m_{1}x$,where $m_{1} = \frac{y}{x}$.
If this line makes an angle $\theta$ with the line $y=mx+c$,then the angle $\theta$ between them is given by the formula:
$\tan \theta = \left| \frac{m_{1}-m}{1+m_{1}m} \right|$
Substituting $m_{1} = \frac{y}{x}$:
$\tan \theta = \left| \frac{\frac{y}{x}-m}{1+\frac{y}{x}m} \right|$
Removing the modulus gives two cases:
Case $I$: $\tan \theta = \frac{\frac{y}{x}-m}{1+\frac{y}{x}m}$
$\Rightarrow \tan \theta + \frac{y}{x}m \tan \theta = \frac{y}{x} - m$
$\Rightarrow m + \tan \theta = \frac{y}{x}(1 - m \tan \theta)$
$\Rightarrow \frac{y}{x} = \frac{m + \tan \theta}{1 - m \tan \theta}$
Case $II$: $-\tan \theta = \frac{\frac{y}{x}-m}{1+\frac{y}{x}m}$
$\Rightarrow -\tan \theta - \frac{y}{x}m \tan \theta = \frac{y}{x} - m$
$\Rightarrow m - \tan \theta = \frac{y}{x}(1 + m \tan \theta)$
$\Rightarrow \frac{y}{x} = \frac{m - \tan \theta}{1 + m \tan \theta}$
Combining both cases,the required equation is $\frac{y}{x} = \frac{m \pm \tan \theta}{1 \mp m \tan \theta}$.

(A) The equations of the given lines are:
$y=3x+1$ .... $(1)$
$2y=x+3$ or $y=\frac{1}{2}x+\frac{3}{2}$ .... $(2)$
$y=mx+4$ .... $(3)$
The slopes of these lines are $m_1=3$,$m_2=\frac{1}{2}$,and $m_3=m$.
Since lines $(1)$ and $(2)$ are equally inclined to line $(3)$,the angle between $(1)$ and $(3)$ is equal to the angle between $(2)$ and $(3)$.
Using the formula for the angle between two lines,$\tan \theta = \left| \frac{m_a - m_b}{1 + m_a m_b} \right|$,we have:
$\left| \frac{3-m}{1+3m} \right| = \left| \frac{\frac{1}{2}-m}{1+\frac{1}{2}m} \right|$
$\left| \frac{3-m}{1+3m} \right| = \left| \frac{1-2m}{2+m} \right|$
Case $1$: $\frac{3-m}{1+3m} = \frac{1-2m}{2+m}$
$(3-m)(2+m) = (1-2m)(1+3m)$
$6 + 3m - 2m - m^2 = 1 + 3m - 2m - 6m^2$
$5m^2 + 5 = 0 \Rightarrow m^2 = -1$ (No real solution).
Case $2$: $\frac{3-m}{1+3m} = -\left( \frac{1-2m}{2+m} \right)$
$(3-m)(2+m) = -(1-2m)(1+3m)$
$6 + m - m^2 = -(1 + m - 6m^2)$
$6 + m - m^2 = -1 - m + 6m^2$
$7m^2 - 2m - 7 = 0$
Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$m = \frac{2 \pm \sqrt{(-2)^2 - 4(7)(-7)}}{2(7)} = \frac{2 \pm \sqrt{4 + 196}}{14} = \frac{2 \pm \sqrt{200}}{14} = \frac{2 \pm 10\sqrt{2}}{14} = \frac{1 \pm 5\sqrt{2}}{7}$.

(A) Let the slope of the required line be $m$. The line passes through $(1, 3)$,so its equation is $y - 3 = m(x - 1)$,or $y = mx + (3 - m)$.
The given line is $y = 3\sqrt{2}x - 1$,which has a slope $m_1 = 3\sqrt{2}$.
The angle $\theta$ between the two lines is given by $\tan \theta = \left| \frac{m - m_1}{1 + m \cdot m_1} \right|$.
Given $\tan \theta = \sqrt{2}$,we have $\sqrt{2} = \left| \frac{m - 3\sqrt{2}}{1 + 3\sqrt{2}m} \right|$.
Case $1$: $\frac{m - 3\sqrt{2}}{1 + 3\sqrt{2}m} = \sqrt{2}$
$m - 3\sqrt{2} = \sqrt{2} + 6m$
$-5m = 4\sqrt{2} \implies m = -\frac{4\sqrt{2}}{5}$.
Case $2$: $\frac{m - 3\sqrt{2}}{1 + 3\sqrt{2}m} = -\sqrt{2}$
$m - 3\sqrt{2} = -\sqrt{2} - 6m$
$7m = 2\sqrt{2} \implies m = \frac{2\sqrt{2}}{7}$.
Using $m = -\frac{4\sqrt{2}}{5}$ in $y - 3 = m(x - 1)$:
$y - 3 = -\frac{4\sqrt{2}}{5}(x - 1)$
$5y - 15 = -4\sqrt{2}x + 4\sqrt{2}$
$4\sqrt{2}x + 5y - (15 + 4\sqrt{2}) = 0$.

(C) Let the points $A$ and $A'$ be $(x, 2)$. The origin is $O(0, 0)$ and $B$ is $(2, 3)$.
Let $m_1$ be the slope of $OB$ and $m_2$ be the slope of $OA$ (or $OA'$).
The slope of $OB$ is $m_1 = \frac{3-0}{2-0} = \frac{3}{2}$.
The slope of $OA$ is $m_2 = \frac{2-0}{x-0} = \frac{2}{x}$.
The angle between $OB$ and $OA$ is given as $\frac{\pi}{4}$.
Using the formula $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$,we have:
$\tan \frac{\pi}{4} = \left| \frac{\frac{3}{2} - \frac{2}{x}}{1 + (\frac{3}{2})(\frac{2}{x})} \right| = 1$
$\left| \frac{\frac{3x - 4}{2x}}{\frac{2x + 6}{2x}} \right| = 1 \Rightarrow \left| \frac{3x - 4}{2x + 6} \right| = 1$
Case $1$: $\frac{3x - 4}{2x + 6} = 1$ $\Rightarrow 3x - 4 = 2x + 6$ $\Rightarrow x = 10$.
Case $2$: $\frac{3x - 4}{2x + 6} = -1$ $\Rightarrow 3x - 4 = -2x - 6$ $\Rightarrow 5x = -2$ $\Rightarrow x = -\frac{2}{5}$.
The points are $A(10, 2)$ and $A'(-\frac{2}{5}, 2)$.
The distance $AA' = |10 - (-\frac{2}{5})| = |10 + \frac{2}{5}| = \frac{52}{5}$.

(D) The line $BC$ has the equation $y = x + 4$,so its slope is $m_1 = 1$. Let the slopes of lines $AB$ and $AC$ be $m_2$ and $m_3$ respectively. The angle between $AB$ and $BC$ is $\frac{\pi}{6}$,so $\tan(\frac{\pi}{6}) = |\frac{m_2 - 1}{1 + m_2}| = \frac{1}{\sqrt{3}}$. This gives $m_2 = 2 \pm \sqrt{3}$.
The line $AB$ passes through $A(1, 2)$ with slope $m_2$. Its equation is $y - 2 = m_2(x - 1)$.
For $m_2 = 2 + \sqrt{3}$,$y - 2 = (2 + \sqrt{3})(x - 1)$. Solving with $y = x + 4$ gives $x + 2 = (2 + \sqrt{3})x - 2 - \sqrt{3}$,so $x = \frac{4 + \sqrt{3}}{1 + \sqrt{3}} = \frac{(4 + \sqrt{3})(\sqrt{3} - 1)}{2} = \frac{4\sqrt{3} - 4 + 3 - \sqrt{3}}{2} = \frac{3\sqrt{3} - 1}{2}$.
For $m_2 = 2 - \sqrt{3}$,$y - 2 = (2 - \sqrt{3})(x - 1)$. Solving with $y = x + 4$ gives $x + 2 = (2 - \sqrt{3})x - 2 + \sqrt{3}$,so $x = \frac{4 - \sqrt{3}}{1 - \sqrt{3}} = \frac{(4 - \sqrt{3})(1 + \sqrt{3})}{-2} = \frac{4 + 4\sqrt{3} - \sqrt{3} - 3}{-2} = \frac{1 + 3\sqrt{3}}{-2}$.
Thus,$\alpha$ and $\gamma$ are $\frac{3\sqrt{3} - 1}{2}$ and $\frac{-(3\sqrt{3} + 1)}{2}$.
$\alpha^2 + \gamma^2 = \frac{27 + 1 - 6\sqrt{3}}{4} + \frac{27 + 1 + 6\sqrt{3}}{4} = \frac{28 + 28}{4} = 14$.

(D) Let $O$ be the origin $(0, 0)$,$A = (5, 2)$,and $B = (2, a)$.
The slope of $OA$ is $m_1 = \frac{2-0}{5-0} = \frac{2}{5}$.
The slope of $OB$ is $m_2 = \frac{a-0}{2-0} = \frac{a}{2}$.
The angle $\theta$ between $OA$ and $OB$ is given as $\frac{\pi}{4}$.
Using the formula $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$,we have:
$\tan \frac{\pi}{4} = \left| \frac{\frac{2}{5} - \frac{a}{2}}{1 + (\frac{2}{5})(\frac{a}{2})} \right|$
$1 = \left| \frac{\frac{4-5a}{10}}{1 + \frac{a}{5}} \right| = \left| \frac{4-5a}{10+2a} \right|$
This implies $4-5a = \pm(10+2a)$.
Case $1$: $4-5a = 10+2a$ $\Rightarrow -7a = 6$ $\Rightarrow a = -\frac{6}{7}$.
Case $2$: $4-5a = -(10+2a)$ $\Rightarrow 4-5a = -10-2a$ $\Rightarrow 3a = 14$ $\Rightarrow a = \frac{14}{3}$.
The product of the possible values of $a$ is $(-\frac{6}{7}) \times (\frac{14}{3}) = -4$.
The absolute value of the product is $|-4| = 4$.

(C) Let the slopes of the two equal sides be $m_1$ and $m_2$.
From the equations $-x+2y=4$ and $x+y=4$,we get $m_1 = \frac{1}{2}$ and $m_2 = -1$.
Let $m$ be the slope of the third side. Since the triangle is isosceles,the angle between the third side and the first side must be equal to the angle between the third side and the second side.
Thus,$\left| \frac{m - m_1}{1 + m \cdot m_1} \right| = \left| \frac{m - m_2}{1 + m \cdot m_2} \right|$.
Substituting the values,$\left| \frac{m - 1/2}{1 + m/2} \right| = \left| \frac{m - (-1)}{1 + m(-1)} \right|$.
$\left| \frac{2m - 1}{2 + m} \right| = \left| \frac{m + 1}{1 - m} \right|$.
This gives two cases:
Case $1$: $\frac{2m - 1}{2 + m} = \frac{m + 1}{1 - m} \implies (2m - 1)(1 - m) = (m + 1)(2 + m) \implies 2m - 2m^2 - 1 + m = 2m + m^2 + 2 + m \implies 3m^2 = -3 \implies m^2 = -1$ (No real solution).
Case $2$: $\frac{2m - 1}{2 + m} = -\left( \frac{m + 1}{1 - m} \right) = \frac{m + 1}{m - 1} \implies (2m - 1)(m - 1) = (m + 1)(m + 2) \implies 2m^2 - 3m + 1 = m^2 + 3m + 2 \implies m^2 - 6m - 1 = 0$.
The sum of the roots of this quadratic equation is given by $-\frac{b}{a} = -\frac{-6}{1} = 6$.

(D) The line passing through the origin and making equal angles with the positive coordinate axes is $y = x$.
$L_1: 2x + y + 6 = 0$ and $L_2: 4x + 2y - p = 0$ (or $2x + y - \frac{p}{2} = 0$).
Since $L_1$ and $L_2$ have the same slope $m = -2$,they are parallel lines.
The line $y = x$ intersects $L_1$ at $A$ and $L_2$ at $B$.
In the right-angled triangle $\triangle AMB$,the angle $\theta$ between the line $y = x$ (slope $m_1 = 1$) and the line $L_2$ (slope $m_2 = -2$) is given by:
$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{1 - (-2)}{1 + (1)(-2)} \right| = \left| \frac{3}{-1} \right| = 3$.
In $\triangle AMB$,$\tan \theta = \frac{AM}{BM}$.
Therefore,$\frac{AM}{BM} = 3$.

(A) Let the slope of the given line $2x - 3y = 5$ be $m_1$. Rewriting the equation as $3y = 2x - 5$,we get $y = \frac{2}{3}x - \frac{5}{3}$. Thus,$m_1 = \frac{2}{3}$.
Let the slope of the required lines be $m$. The angle $\theta$ between the lines is $45^{\circ}$.
Using the formula $\tan \theta = \left| \frac{m - m_1}{1 + m \cdot m_1} \right|$,we have:
$\tan 45^{\circ} = \left| \frac{m - \frac{2}{3}}{1 + m \cdot \frac{2}{3}} \right|$
$1 = \left| \frac{3m - 2}{3 + 2m} \right|$
This gives two cases:
Case $1$: $\frac{3m - 2}{3 + 2m} = 1$ $\Rightarrow 3m - 2 = 3 + 2m$ $\Rightarrow m = 5$.
Case $2$: $\frac{3m - 2}{3 + 2m} = -1$ $\Rightarrow 3m - 2 = -3 - 2m$ $\Rightarrow 5m = -1$ $\Rightarrow m = \frac{-1}{5}$.
Therefore,the slopes are $5$ and $\frac{-1}{5}$.

(B) The slope of the line $4x - 2y + 13 = 0$ is $m_1 = -\frac{4}{-2} = 2$.
The line making equal intercepts with the coordinate axes has the form $\frac{x}{a} + \frac{y}{a} = 1$,which simplifies to $x + y = a$. The slope of this line is $m_2 = -1$.
The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values: $\tan \theta = \left| \frac{2 - (-1)}{1 + (2)(-1)} \right| = \left| \frac{3}{1 - 2} \right| = \left| \frac{3}{-1} \right| = 3$.
Therefore,$\theta = \tan^{-1}(3)$.

(B) The given equations of the lines are $x \cos 30^{\circ} + y \sin 30^{\circ} = 3$ and $x \cos 60^{\circ} + y \sin 60^{\circ} = 5$.
These are in the normal form $x \cos \alpha + y \sin \alpha = p$,where $\alpha$ is the angle made by the normal to the line with the positive $x$-axis.
The angle of the normal to the first line is $\alpha_1 = 30^{\circ}$ and to the second line is $\alpha_2 = 60^{\circ}$.
The angle $\theta$ between two lines is equal to the angle between their normals.
Therefore,$\theta = |\alpha_2 - \alpha_1| = |60^{\circ} - 30^{\circ}| = 30^{\circ}$.

(A) Let $m_1$ be the slope of the line passing through $(-2, 6)$ and $(4, 8)$.
$m_1 = \frac{8 - 6}{4 - (-2)} = \frac{2}{6} = \frac{1}{3}$.
Let $m_2$ be the slope of the line passing through $(8, 12)$ and $(x, 24)$.
$m_2 = \frac{24 - 12}{x - 8} = \frac{12}{x - 8}$.
Since the lines are perpendicular,$m_1 \times m_2 = -1$.
$\frac{1}{3} \times \frac{12}{x - 8} = -1$.
$\frac{4}{x - 8} = -1$.
$4 = -(x - 8)$.
$4 = -x + 8$.
$x = 8 - 4 = 4$.

(A) Given $\theta = \frac{\pi}{4} = 45^{\circ}$ and $m_1 = \frac{1}{2}$.
Using the formula for the angle between two lines:
$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$
$\tan 45^{\circ} = \left| \frac{\frac{1}{2} - m_2}{1 + \frac{1}{2} m_2} \right|$
$1 = \left| \frac{1 - 2m_2}{2 + m_2} \right|$
This implies $\frac{1 - 2m_2}{2 + m_2} = 1$ or $\frac{1 - 2m_2}{2 + m_2} = -1$.
Case $1$: $1 - 2m_2 = 2 + m_2$ $\Rightarrow -3m_2 = 1$ $\Rightarrow m_2 = -\frac{1}{3}$.
Case $2$: $1 - 2m_2 = -(2 + m_2)$ $\Rightarrow 1 - 2m_2 = -2 - m_2$ $\Rightarrow m_2 = 3$.
Thus,the slope of the other line is $3$ or $-\frac{1}{3}$.

(A) The equation of the first line is $x \sin \theta - y \cos \theta = 5$. Its slope $m_1 = \frac{\sin \theta}{\cos \theta} = \tan \theta$.
The equation of the second line is $x \sin \alpha - y \cos \alpha + 11 = 0$. Its slope $m_2 = \frac{\sin \alpha}{\cos \alpha} = \tan \alpha$.
Let $\beta$ be the acute angle between the lines.
Then,$\tan \beta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\tan \theta - \tan \alpha}{1 + \tan \theta \tan \alpha} \right|$.
Using the trigonometric identity $\tan(\theta - \alpha) = \frac{\tan \theta - \tan \alpha}{1 + \tan \theta \tan \alpha}$,we get $\tan \beta = |\tan(\theta - \alpha)|$.
Therefore,$\beta = |\theta - \alpha|$.

(D) The given equations of the lines are $y = \sqrt{3}x - 1$ and $y = \frac{1}{\sqrt{3}}x - \frac{7}{\sqrt{3}}$.
Comparing these with $y = mx + c$,the slopes are $m_{1} = \sqrt{3}$ and $m_{2} = \frac{1}{\sqrt{3}}$.
The angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{m_{1} - m_{2}}{1 + m_{1}m_{2}} \right|$.
Substituting the values: $\tan \theta = \left| \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + (\sqrt{3})(\frac{1}{\sqrt{3}})} \right| = \left| \frac{\frac{3-1}{\sqrt{3}}}{1 + 1} \right| = \left| \frac{2/\sqrt{3}}{2} \right| = \frac{1}{\sqrt{3}}$.
Since $\tan \theta = \frac{1}{\sqrt{3}}$,we have $\theta = 30^{\circ}$.

(D) The given lines are $x-3=0$ $(i)$ and $x+y=19$ (ii).
For line $(i)$,$x=3$,which is a vertical line parallel to the $y$-axis. The angle it makes with the positive $x$-axis is $\theta_1 = 90^{\circ}$.
For line (ii),$x+y=19$,we can rewrite it as $y = -x + 19$. Comparing this with the slope-intercept form $y = mx + c$,the slope $m_2 = -1$.
Since $m_2 = \tan \theta_2 = -1$,the angle $\theta_2 = 135^{\circ}$.
The angle between the two lines is $|\theta_2 - \theta_1| = |135^{\circ} - 90^{\circ}| = 45^{\circ}$.
Since $45^{\circ}$ is an acute angle,the required angle is $45^{\circ}$.

(D) Given equations of lines:
$x+(a-1)y=1$ and $2x+a^2y=1$.
Slope of $x+(a-1)y=1$ is $m_1 = \frac{-1}{a-1}$.
Slope of $2x+a^2y=1$ is $m_2 = \frac{-2}{a^2}$.
Since the lines are perpendicular,$m_1 \times m_2 = -1$.
$\frac{-1}{a-1} \times \frac{-2}{a^2} = -1$
$\frac{2}{a^2(a-1)} = -1$
$2 = -a^3 + a^2$
$a^3 - a^2 + 2 = 0$.
Factoring the cubic equation: $(a+1)(a^2-2a+2) = 0$.
Since $a^2-2a+2 = (a-1)^2 + 1 > 0$ for all real $a$,we must have $a+1=0$,so $a=-1$.
Substituting $a=-1$ into the line equations:
$x+(-1-1)y=1 \Rightarrow x-2y=1$.
$2x+(-1)^2y=1 \Rightarrow 2x+y=1$.
Solving the system:
$x-2y=1$ $(i)$
$2x+y=1$ (ii)
Multiply (ii) by $2$: $4x+2y=2$.
Adding $(i)$ and (ii): $5x=3 \Rightarrow x=\frac{3}{5}$.
Substituting $x=\frac{3}{5}$ into (ii): $2(\frac{3}{5})+y=1 \Rightarrow y=1-\frac{6}{5} = -\frac{1}{5}$.
The point of intersection is $(\frac{3}{5}, -\frac{1}{5})$.
The distance from the origin $(0,0)$ is $\sqrt{(\frac{3}{5})^2 + (-\frac{1}{5})^2} = \sqrt{\frac{9}{25} + \frac{1}{25}} = \sqrt{\frac{10}{25}} = \sqrt{\frac{2}{5}}$.

(D) Let the given line be $L: 3x + y + 4 = 0$. The slope of $L$ is $m = -3$.
Let the slopes of lines $L_1$ and $L_2$ be $m_1 = \frac{3}{4}$ and $m_2$ respectively.
Since the distances measured along $L_1$ and $L_2$ from point $A(2, 5)$ to the line $L$ are equal,the angles $\theta_1$ and $\theta_2$ that $L_1$ and $L_2$ make with the line $L$ must satisfy $\tan \theta_1 = \tan \theta_2$ or $\tan \theta_1 = -\tan \theta_2$.
Using the formula for the angle between two lines with slopes $m$ and $m'$,$\tan \theta = \left| \frac{m - m'}{1 + mm'} \right|$.
For $L_1$: $\tan \theta = \left| \frac{-3 - 3/4}{1 + (-3)(3/4)} \right| = \left| \frac{-15/4}{1 - 9/4} \right| = \left| \frac{-15/4}{-5/4} \right| = 3$.
For $L_2$: $\left| \frac{-3 - m_2}{1 - 3m_2} \right| = 3$.
Case $1$: $\frac{-3 - m_2}{1 - 3m_2} = 3$ $\Rightarrow -3 - m_2 = 3 - 9m_2$ $\Rightarrow 8m_2 = 6$ $\Rightarrow m_2 = \frac{3}{4}$ (This is $L_1$ itself).
Case $2$: $\frac{-3 - m_2}{1 - 3m_2} = -3$ $\Rightarrow -3 - m_2 = -3 + 9m_2$ $\Rightarrow 10m_2 = 0$ $\Rightarrow m_2 = 0$.
Thus,the slope of $L_2$ is $0$.

(B) The equation of a straight line passing through $(3,-2)$ with slope $m$ is given by $y+2=m(x-3)$ $(i)$.
The given line is $\sqrt{3} x+y=1$,which can be written as $y=-\sqrt{3} x+1$. The slope of this line is $m_1=-\sqrt{3}$.
The angle between the two lines is $60^{\circ}$. Using the formula $\tan \theta = \left| \frac{m-m_1}{1+m m_1} \right|$,we have $\tan 60^{\circ} = \left| \frac{m-(-\sqrt{3})}{1+m(-\sqrt{3})} \right|$.
$\sqrt{3} = \left| \frac{m+\sqrt{3}}{1-\sqrt{3} m} \right|$.
Case $1$: $\sqrt{3} = \frac{m+\sqrt{3}}{1-\sqrt{3} m} \Rightarrow \sqrt{3} - 3m = m + \sqrt{3} \Rightarrow 4m = 0 \Rightarrow m = 0$.
Substituting $m=0$ into $(i)$,we get $y+2=0(x-3) \Rightarrow y+2=0$. This line is parallel to the $X$-axis and does not intersect the $X$-axis (unless it is the $X$-axis itself,which it is not).
Case $2$: $-\sqrt{3} = \frac{m+\sqrt{3}}{1-\sqrt{3} m} \Rightarrow -\sqrt{3} + 3m = m + \sqrt{3} \Rightarrow 2m = 2\sqrt{3} \Rightarrow m = \sqrt{3}$.
Substituting $m=\sqrt{3}$ into $(i)$,we get $y+2=\sqrt{3}(x-3) \Rightarrow y+2=\sqrt{3}x-3\sqrt{3} \Rightarrow y-\sqrt{3}x+2+3\sqrt{3}=0$.

(C) Let the slope of the required line be $m$. The slope of the given line $x-2y-3=0$ is $m_1 = \frac{1}{2}$.
Since the angle between the lines is $45^{\circ}$,we have $\tan 45^{\circ} = \left| \frac{m - m_1}{1 + m \cdot m_1} \right|$.
$1 = \left| \frac{m - 1/2}{1 + m/2} \right| = \left| \frac{2m - 1}{2 + m} \right|$.
This gives two cases: $\frac{2m - 1}{2 + m} = 1$ or $\frac{2m - 1}{2 + m} = -1$.
Case $1$: $2m - 1 = 2 + m \Rightarrow m = 3$.
The equation of the line is $y - 2 = 3(x - 3) \Rightarrow 3x - y - 7 = 0$.
Case $2$: $2m - 1 = -2 - m \Rightarrow 3m = -1 \Rightarrow m = -1/3$.
The equation of the line is $y - 2 = -1/3(x - 3) \Rightarrow 3y - 6 = -x + 3 \Rightarrow x + 3y - 9 = 0$.
Thus,the equations are $3x - y - 7 = 0$ and $x + 3y - 9 = 0$.

(C) The slope of the line $x+y=3$ is $m_1 = -1$.
The slope of the line joining $(1,1)$ and $(-3,4)$ is $m_2 = \frac{4-1}{-3-1} = \frac{3}{-4} = -\frac{3}{4}$.
The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by $\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$.
Substituting the values: $\tan \theta = \left| \frac{-\frac{3}{4} - (-1)}{1 + (-1)(-\frac{3}{4})} \right| = \left| \frac{-\frac{3}{4} + 1}{1 + \frac{3}{4}} \right| = \left| \frac{\frac{1}{4}}{\frac{7}{4}} \right| = \frac{1}{7}$.
Therefore,$\theta = \tan ^{-1}\left(\frac{1}{7}\right)$.

(C) The given equations of the lines are:
$2x + 3y - 3 = 0$ --- $(1)$
$x + ky + 7 = 0$ --- $(2)$
The slope of line $(1)$ is $m_1 = -\frac{2}{3}$.
The slope of line $(2)$ is $m_2 = -\frac{1}{k}$.
Since the lines are perpendicular,the product of their slopes must be $-1$:
$m_1 \times m_2 = -1$
$(-\frac{2}{3}) \times (-\frac{1}{k}) = -1$
$\frac{2}{3k} = -1$
$2 = -3k$
$k = -\frac{2}{3}$

(A) The equation of the first line is $lx + my = n$,which can be written as $y = -\frac{l}{m}x + \frac{n}{m}$. The slope $m_1 = -\frac{l}{m}$.
The equation of the second line is $l'x + m'y = n'$,which can be written as $y = -\frac{l'}{m'}x + \frac{n'}{m'}$. The slope $m_2 = -\frac{l'}{m'}$.
Two lines are perpendicular if the product of their slopes is $-1$,i.e.,$m_1 \times m_2 = -1$.
$\left(-\frac{l}{m}\right) \times \left(-\frac{l'}{m'}\right) = -1$
$\frac{ll'}{mm'} = -1$
$ll' = -mm'$
$ll' + mm' = 0$.

(B) The slope of diagonal $AC$ is $m_{AC} = \frac{-1-3}{1-(-3)} = \frac{-4}{4} = -1$.
The slope of diagonal $BD$ is $m_{BD} = \frac{-2-1}{-2-1} = \frac{-3}{-3} = 1$.
Since the product of the slopes is $m_{AC} \times m_{BD} = -1 \times 1 = -1$,the diagonals $AC$ and $BD$ are perpendicular to each other.
Therefore,the angle between the diagonals $AC$ and $BD$ is $\frac{\pi}{2}$.

(A) Given,the equations of the straight lines are:
$l_1: 3x + 4y + 9 = 0$ ...$(i)$
$l_2: x - 7y - 22 = 0$ ...(ii)
The slope of line $l_1$ is $m_1 = -\frac{3}{4}$.
The slope of line $l_2$ is $m_2 = -\frac{1}{-7} = \frac{1}{7}$.
Let $\theta$ be the angle between the lines $l_1$ and $l_2$. The formula for the angle between two lines is:
$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$
Substituting the values of $m_1$ and $m_2$:
$\tan \theta = \left| \frac{-\frac{3}{4} - \frac{1}{7}}{1 + (-\frac{3}{4})(\frac{1}{7})} \right|$
$\tan \theta = \left| \frac{-\frac{21+4}{28}}{1 - \frac{3}{28}} \right| = \left| \frac{-\frac{25}{28}}{\frac{25}{28}} \right|$
$\tan \theta = |-1| = 1$
Since $\tan \theta = 1$,we have $\theta = \frac{\pi}{4}$.

(C) The given lines are $2x + 11y - 7 = 0$ and $x + 3y + 5 = 0$.
The slopes of these lines are $m_1 = -\frac{2}{11}$ and $m_2 = -\frac{1}{3}$.
The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by $\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$.
Substituting the values: $\tan \theta = \left|\frac{-\frac{2}{11} - (-\frac{1}{3})}{1 + (-\frac{2}{11})(-\frac{1}{3})}\right| = \left|\frac{-\frac{6}{33} + \frac{11}{33}}{1 + \frac{2}{33}}\right| = \left|\frac{\frac{5}{33}}{\frac{35}{33}}\right| = \frac{5}{35} = \frac{1}{7}$.
Therefore,$\theta = \tan^{-1}\left(\frac{1}{7}\right)$.

(B) The given lines are $L_1: x-y=0$ and $L_2: y=0$.
Comparing $L_1$ with $y=m_1x+c_1$,we get $m_1 = 1$.
Comparing $L_2$ with $y=m_2x+c_2$,we get $m_2 = 0$.
The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values,$\tan \theta = \left| \frac{1 - 0}{1 + (1)(0)} \right| = 1$.
Therefore,$\theta = \tan^{-1}(1) = \frac{\pi}{4}$.

(D) The given lines are $y=3x+1$ (slope $m_1=3$) and $2y=x+3$ or $y=\frac{1}{2}x+\frac{3}{2}$ (slope $m_2=\frac{1}{2}$).
Let the third line be $y=mx+4$ with slope $m$.
Since the lines are equally inclined to the third line,the angle between the first and third line equals the angle between the second and third line:
$\left|\frac{m_1-m}{1+m_1m}\right| = \left|\frac{m_2-m}{1+m_2m}\right|$
Substituting the slopes:
$\left|\frac{3-m}{1+3m}\right| = \left|\frac{\frac{1}{2}-m}{1+\frac{1}{2}m}\right| = \left|\frac{1-2m}{2+m}\right|$
Case $1$: $\frac{3-m}{1+3m} = \frac{1-2m}{2+m}$
$(3-m)(2+m) = (1-2m)(1+3m)$
$6+m-m^2 = 1+m-6m^2$
$5m^2+5 = 0 \Rightarrow m^2 = -1$ (No real solution).
Case $2$: $\frac{3-m}{1+3m} = -\left(\frac{1-2m}{2+m}\right)$
$(3-m)(2+m) = -(1-2m)(1+3m)$
$6+m-m^2 = -(1+m-6m^2) = -1-m+6m^2$
$7m^2-2m-7 = 0$
Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:
$m = \frac{2 \pm \sqrt{4-4(7)(-7)}}{14} = \frac{2 \pm \sqrt{4+196}}{14} = \frac{2 \pm \sqrt{200}}{14} = \frac{2 \pm 10\sqrt{2}}{14} = \frac{1 \pm 5\sqrt{2}}{7}$.

(A) Let the angle between the two lines $S_1$ and $S_2$ be $\theta$.
When $S_1$ is reflected in $S_2$,the angle between the reflected line $S_1'$ and $S_2$ is $\theta$.
When $S_2$ is reflected in $S_1$,the angle between the reflected line $S_2'$ and $S_1$ is $\theta$.
For the reflected lines $S_1'$ and $S_2'$ to coincide,the total angle covered by the lines must be $\pi$ radians.
Specifically,the angle between $S_1'$ and $S_2'$ is $3\theta = \pi$.
Therefore,$\theta = \frac{\pi}{3}$.

(A) Let the slope of the lines inclined at an angle of $60^{\circ}$ with line $L$ be $n$.
Given the slope of line $L$ is $m=1$.
The angle $\theta$ between two lines with slopes $m$ and $n$ is given by $\tan \theta = \left| \frac{n-m}{1+nm} \right|$.
Substituting the values,we have $\tan 60^{\circ} = \left| \frac{n-1}{1+n} \right| = \sqrt{3}$.
Squaring both sides,we get $\frac{(n-1)^2}{(n+1)^2} = 3$.
$(n-1)^2 = 3(n+1)^2
$ $\Rightarrow n^2 - 2n + 1 = 3(n^2 + 2n + 1)
$ $\Rightarrow n^2 - 2n + 1 = 3n^2 + 6n + 3
$ $\Rightarrow 2n^2 + 8n + 2 = 0
$ $\Rightarrow n^2 + 4n + 1 = 0$.
The roots of this quadratic equation represent the slopes of the two lines.
The product of the roots (slopes) is given by the constant term divided by the coefficient of $n^2$,which is $\frac{1}{1} = 1$.

(A) Let the slope of line $L$ be $m$. Since the two given lines are perpendicular to line $L$,they must be parallel to each other.
Let the slopes of the two lines be $m_1$ and $m_2$ respectively.
For the first line $p(p^2+1)x - y + q = 0$,the slope $m_1 = p(p^2+1)$.
For the second line $(p^2+1)^2 x + (p^2+1)y + 2q = 0$,the slope $m_2 = -\frac{(p^2+1)^2}{(p^2+1)} = -(p^2+1)$.
Since the lines are parallel,$m_1 = m_2$:
$p(p^2+1) = -(p^2+1)$.
Since $(p^2+1) \neq 0$ for any real $p$,we can divide by $(p^2+1)$:
$p = -1$.
Thus,there is exactly one value of $p$.

(A) Let $m_1$ be the slope of the line passing through $(3,2)$ and $m_2$ be the slope of $x-2y=3$.
$m_2 = \frac{-1}{-2} = \frac{1}{2}$.
The angle between the lines is $45^{\circ}$,so $\tan 45^{\circ} = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
$1 = \left| \frac{m_1 - 1/2}{1 + m_1/2} \right| = \left| \frac{2m_1 - 1}{2 + m_1} \right|$.
Case $1$: $\frac{2m_1 - 1}{2 + m_1} = 1$ $\Rightarrow 2m_1 - 1 = 2 + m_1$ $\Rightarrow m_1 = 3$.
The equation is $y - 2 = 3(x - 3)$ $\Rightarrow y - 2 = 3x - 9$ $\Rightarrow 3x - y = 7$.
Case $2$: $\frac{2m_1 - 1}{2 + m_1} = -1$ $\Rightarrow 2m_1 - 1 = -2 - m_1$ $\Rightarrow 3m_1 = -1$ $\Rightarrow m_1 = -1/3$.
The equation is $y - 2 = -\frac{1}{3}(x - 3)$ $\Rightarrow 3y - 6 = -x + 3$ $\Rightarrow x + 3y = 9$.
Thus,the equations are $3x - y = 7$ and $x + 3y = 9$.

(B) The given line is $(2x + 3y + 4) + \lambda(6x - y + 12) = 0$.
Rearranging the terms,we get $(2 + 6\lambda)x + (3 - \lambda)y + (4 + 12\lambda) = 0$.
The slope of this line $(m_1)$ is $-\frac{2 + 6\lambda}{3 - \lambda}$.
The second line is $7x + 5y = 2$,which can be written as $5y = -7x + 2$,or $y = -\frac{7}{5}x + \frac{2}{5}$.
The slope of this line $(m_2)$ is $-\frac{7}{5}$.
Since the lines are perpendicular,the product of their slopes is $-1$,so $m_1 \times m_2 = -1$.
$(-\frac{2 + 6\lambda}{3 - \lambda}) \times (-\frac{7}{5}) = -1$.
$\frac{14 + 42\lambda}{5(3 - \lambda)} = -1$.
$14 + 42\lambda = -5(3 - \lambda)$.
$14 + 42\lambda = -15 + 5\lambda$.
$42\lambda - 5\lambda = -15 - 14$.
$37\lambda = -29$.
$\lambda = -\frac{29}{37}$.

(C) The slope of the given line $2x-y+3=0$ is $m_1 = 2$.
Let the slope of line $L$ be $m$.
The angle between the lines is $60^{\circ}$,so $\tan 60^{\circ} = |\frac{m-m_1}{1+m m_1}|$.
$\sqrt{3} = |\frac{m-2}{1+2m}|$.
This gives two cases:
Case $1$: $\frac{m-2}{1+2m} = \sqrt{3} \Rightarrow m-2 = \sqrt{3} + 2\sqrt{3}m \Rightarrow m(1-2\sqrt{3}) = 2+\sqrt{3} \Rightarrow m = \frac{2+\sqrt{3}}{1-2\sqrt{3}} = \frac{(2+\sqrt{3})(1+2\sqrt{3})}{1-12} = \frac{2+4\sqrt{3}+\sqrt{3}+6}{-11} = \frac{8+5\sqrt{3}}{-11}$.
Case $2$: $\frac{m-2}{1+2m} = -\sqrt{3} \Rightarrow m-2 = -\sqrt{3} - 2\sqrt{3}m \Rightarrow m(1+2\sqrt{3}) = 2-\sqrt{3} \Rightarrow m = \frac{2-\sqrt{3}}{1+2\sqrt{3}} = \frac{(2-\sqrt{3})(1-2\sqrt{3})}{1-12} = \frac{2-4\sqrt{3}-\sqrt{3}+6}{-11} = \frac{8-5\sqrt{3}}{-11}$.
Since $L$ makes an acute angle with the positive $X$-axis,$m > 0$.
Comparing the two slopes,$\frac{8-5\sqrt{3}}{-11} = \frac{5\sqrt{3}-8}{11} \approx \frac{5(1.732)-8}{11} = \frac{8.66-8}{11} > 0$.
Thus,$m = \frac{5\sqrt{3}-8}{11}$.
The equation of line $L$ is $y - 0 = m(x - 2) \Rightarrow y = \frac{5\sqrt{3}-8}{11}(x - 2)$.
The $Y$-intercept is found by setting $x = 0$: $y = \frac{5\sqrt{3}-8}{11}(-2) = \frac{16-10\sqrt{3}}{11}$.